ABCD is a parallelogram. AB = 3√2, AD = 8, ∠BAD = 45◦
, M is a point on CD, CM : MD = 1 : 2. N is a point on AD, AN : ND = 3 : 1. Find:
a) the length of AM;
b) the length of BN;
c) the acute angle between AM and BN

Given the parallelogram \(ABCD\) with the following properties:

- \(AB = 3\sqrt{2}\)
- \(AD = 8\)
- \(\angle BAD = 45^\circ\)
- Point \(M\) divides \(CD\) in the ratio \(CM : MD = 1 : 2\)
- Point \(N\) divides \(AD\) in the ratio \(AN : ND = 3 : 1\)

Let's find:

### a) The Length of \(AM\)

To find \(AM\), we first need to determine the coordinates of points \(C\) and \(D\). Since \(ABCD\) is a parallelogram, we can set \(A\) at \((0, 0)\), \(B\) at \((3\sqrt{2}, 0)\), \(D\) at \((0, 8)\), and \(C\) at \((3\sqrt{2}, 8)\).

**Coordinates of \(M\):**

Point \(M\) divides \(CD\) in the ratio \(1:2\), so using the section formula, the coordinates of \(M\) are:

\[
M = \left(\frac{2 \cdot 3\sqrt{2} + 1 \cdot 3\sqrt{2}}{1+2}, \frac{2 \cdot 8 + 1 \cdot 8}{1+2}\right) = \left(2\sqrt{2}, \frac{24}{3}\right) = (2\sqrt{2}, 8)
\]

**Length of \(AM\):**

\[
AM = \sqrt{(2\sqrt{2} - 0)^2 + (8 - 0)^2} = \sqrt{8 + 64} = \sqrt{72} = 6\sqrt{2}
\]

### b) The Length of \(BN\)

**Coordinates of \(N\):**

Point \(N\) divides \(AD\) in the ratio \(3:1\), so using the section formula, the coordinates of \(N\) are:

\[
N = \left(\frac{3 \cdot 0 + 1 \cdot 0}{3+1}, \frac{3 \cdot 8 + 1 \cdot 0}{3+1}\right) = (0, 6)
\]

**Length of \(BN\):**

\[
BN = \sqrt{(3\sqrt{2} - 0)^2 + (0 - 6)^2} = \sqrt{18 + 36} = \sqrt{54} = 3\sqrt{6}
\]

### c) The Acute Angle Between \(AM\) and \(BN\)

To find the acute angle between vectors \( \overrightarrow{AM} \) and \( \overrightarrow{BN} \), we use the dot product formula:

\[
\cos \theta = \frac{\overrightarrow{AM} \cdot \overrightarrow{BN}}{|\overrightarrow{AM}| \times |\overrightarrow{BN}|}
\]

**Vectors:**

\[
\overrightarrow{AM} = (2\sqrt{2}, 8) \quad \text{and} \quad \overrightarrow{BN} = (-3\sqrt{2}, -6)
\]

**Dot Product:**

\[
\overrightarrow{AM} \cdot \overrightarrow{BN} = (2\sqrt{2})(-3\sqrt{2}) + (8)(-6) = -12 - 48 = -60
\]

**Magnitude of Vectors:**

\[
|\overrightarrow{AM}| = 6\sqrt{2}, \quad |\overrightarrow{BN}| = 3\sqrt{6}
\]

**Cosine of the Angle:**

\[
\cos \theta = \frac{-60}{(6\sqrt{2})(3\sqrt{6})} = \frac{-60}{18\sqrt{12}} = \frac{-60}{36\sqrt{3}} = -\frac{5}{3\sqrt{3}} = -\frac{5\sqrt{3}}{9}
\]

**Angle \( \theta \):**

\[
\theta = \cos^{-1}\left(-\frac{5\sqrt{3}}{9}\right)
\]

Since we need the acute angle, we consider the positive value:

\[
\theta = 180^\circ - \cos^{-1}\left(\frac{5\sqrt{3}}{9}\right)
\]

Would you like further details or calculations? Here are related questions you might find helpful:

1. How do you derive the section formula used here?
2. What is the geometric interpretation of the dot product of two vectors?
3. How would the problem change if \( \angle BAD \) were different?
4. What is the significance of dividing a line segment in a given ratio?
5. How would you find coordinates in 3D using similar methods?

**Tip:** When working with ratios and coordinates, always use the section formula for accurate results!

ABCD is a parallelogram. AB = 3√2, AD = 8, ∠BAD = 45°, M is a point on CD, CM : MD = 1 : 2. N is a point on AD, AN : ND = 3 : 1. Find: a) the length of AM; b) the length of BN; c) the acute angle between AM and BN.

Solve the geometry problem involving a parallelogram ABCD with sides AB = 3√2, AD = 8, and angle BAD = 45°. Learn how to find the lengths of AM, BN, and the acute angle between them using coordinate geometry, section formulas, and trigonometry. Suitable for Grades 10-12.

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